Question

Ryan is under house arrest and can only stay within 0.100 km of his home. Ryan chases a squirrel 85.4 m, 27.2 degrees south of west. The squirrel changes direction and ryan follows closely, travelling 85.7m, 73.0 degrees north of west before the squirrel goes up a tree.

a) when he arrives at the tree, how much is ryan outside of the safe zone? ANSWER: 10

b) If he can run 5.0 m/s on a direct path home, will he make it back whithn the safe zone in under 15 s? ANSWE:NO. 22 s

Answer 1

a)
AB_x = - AB Cos27.2 = - 85.4 Cos27.2

BC_x = - BC Cos27.2 = - 85.7 Cos73

Net displacement in x-direction = X = - 85.4 Cos27.2 - 85.7 Cos73 = - 101.01 m

AB_y = - AB Sin27.2 = - 85.4 Sin27.2

BC_y = BC Cos27.2 = 85.7 Sin73

Net displacement in y-direction = Y = - 85.4 Sin27.2 + 85.7 Sin73 = 42.92 m

Net displacement = sqrt(X² + Y²) = sqrt((-101.01)² + (42.92)²) = 109.75 m

safe distance = 100 m

distance outside = 109.75 - 100 = 9.75 = 10 m

b)

t = 15 sec

v = speed = 5 m/s

distance travelled in 15 sec = D = 5 x 15 = 75 m